Which integral gives the length of the graph of $f(x)=\ln(x^2)$ between $x=a$ and $x=b$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_a^b\sqrt{1+2\ln x~}~dx$ (Choice B) B $ \int_a^b\sqrt{1+(2\ln x)^2}~dx$ (Choice C) C $ \int_a^b\sqrt{1+\dfrac{2}{x}~}~dx$ (Choice D) D $ \int_a^b\sqrt{1+\dfrac{4}{x^2}~}~dx$
Solution: Recall that the formula for arc length of $ ~f(x)~$ over $ ~[a, b]~$ is $ L=\int_a^b\sqrt{1+\big[f~^\prime(x)\big]^2}~dx\,$. First calculate $ f\,^\prime(x)~$ and simplify. $ f ^\prime(x)=\dfrac1{x^2}\cdot2x=\dfrac{2}{x}$ Next, use the formula above to write the integral expression that gives the arc length in question. $ L=\int_{a}^b\sqrt{1+\bigg(\dfrac{2}{x}\bigg)^2}~dx=\int_a^b\sqrt{1+\dfrac{4}{x^2}~}~dx$